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If a space has curvature, then the curvature can be seen intrinsically by finding sums of angles in triangles made of geodesics. Under general relativity, space-time is curved on local scales. On global scales, experiments have determined that space-time is almost totally flat. If one were to, for each of three points in a gravitational gradient (like near a black hole), find the angle between the sight lines to the other two points and then add up all of the angles, would it add up to 180 (degrees) or perhaps it would show intrinsic curvature?

Is the way of thinking about this good, because it is quite general and conceptual. I don't actually know the equations for general relativity, but this is what I thought of when I was thinking about intrinsic geometric properties.

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I think that there is a problem in this view: in GR we are not dealing with Riemannian metrics, but instead with Lorentzian ones. That is, the signture of the metric is "(-,+,+,+)". The notion of "angle", something that varies between 0 and \Pi and denotes the inclination of one direction with respect to the other, is good for Riemannian metrics. There is no sense (as far as I know, please someone correct me if this is not true) in talking about this kind of angle in Lorentzian manifolds. –  Ronaldo Oct 19 '10 at 21:00
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What is well-defined is the "hyperbolic angle" between two time-like vectors, but this is another thing. Maybe you can look at a Riemannian submanifold (with a positive-definite metric) to analyze angles, but I don't know if this helps. –  Ronaldo Oct 19 '10 at 21:01
    
@Ronaldo: Now looking at the comments I think I have just done what you had in mind :) Greets –  Robert Filter Jan 18 '11 at 14:36

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your question tackles some basic concepts of general relativity and hence pseudo-Riemannian manifolds. To answer the several parts your question, we will have to go a little into detail.

Limiting behaviour of general relativity

First of all, I have to confess that you somehow got the limiting behaviour of Einstein's theory of gravitation the other way round.
Locally, you will always be able to construct a Minkowski spacetime with metric $$\eta = -c^2 dt^2 + d\mathbf{r}^2$$ such that your original spacetime metric $$g=\eta + \text{higher order terms}_{ij}dx^i dx^j$$ is kind of Taylor expanded locally. This linearization works pretty well for "normal life" gravitation and you can explain gravitational waves or get Newtonian gravitation. How good a linear approximation can be will tell you the value of the scalar Riemann curvature $R = R^a_{\,a}$.

For long-range interactions, this picture fails as a Taylor expansion will be more and more inaccurate. Then, you have to take the full theory into account and can, under further assumptions, explain things corresponding to cosmology.

Measuring angles

First of all, we have to make a simplification. If you would assume that you want to measure the angles of your triangle on a dynamic spacetime, you would have some conceptual problems defining how to return to the point where you started, and, maybe worse: what is an angle then?

So, it is convenient to assume that you want to do that in a stationary spacetime. That means there exists a local isometry of the metric $\varphi$ and a corresponding Killing field $\xi = \dot{\varphi}(\tau)$ which is properly parametrized, say $g\left( \xi,\xi \right) = -c^2\,\forall\tau$, $$\varphi^\star g = g$$

Now we can define something like an ordinary angle which I guess you had in mind. Consider the metric $$\tilde{g}\left(X,Y\right) = g\left(X,Y\right) - \frac{1}{g(\xi,\xi)}g\left(\xi,Y\right)g\left(\xi,X\right)$$ which only takes the spatial part of the scalar product of the vectors $X$ and $Y$, then you can assign an angle $\theta$ to these quantities in the usual sense as $$\cos \theta = \frac{\tilde{g}\left(X,Y\right)}{\sqrt{\tilde{g}\left(X,X\right)\cdot\tilde{g}\left(Y,Y\right)}}$$

Now you have to think of what $X$ and $Y$ should be. Since you want to compare these quantities at three different points, you will have to parallel transport them between these points using a geodesic motion, again in the 4D setup.
Light geodesics will be totally different from e.g. human ones. So, generally, the angles will depend on how you transport.

So, DarthPickley, if you have a light ray travelling on the event horizon of a Schwarzschild black hole, what would be the sum of the angles of a corresponding "triangle"?

Sincerely

Robert

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This seems a bit overly complicated - it would make a lot more sense to me to define a triangle (not necessarily 'the' triangle!) between three points A, B, and C as being described by chosen geodesics between A and B, between B and C and between C and A (of course, if any of these sets are empty then there is no such triangle!). Stationarity ensures that we can speak only about the spatial portions of the metric and not the temporal portions (which I think was the poster's intention), and then all the quantities are well-defined... –  Steven Stadnicki Jan 18 '11 at 18:45
    
e.g., as the two geodesics meeting at A should each have well-defined tangent vectors at A barring some sort of metric shenanigans, then you can define the angle of the triangle at A; this quantity is scalar and so transports trivially, and so you can add the three angles of the triangle. –  Steven Stadnicki Jan 18 '11 at 18:46
    
@Steven: Thank you for your thoughts - I am not sure if I get them correctly, though. Assuming you are in a dynamic spacetime you can go from A to B to C on some geodesic. How do you go to A from C? It wont be possible with respect to causality. So you will have to define an equivalence class of points A - but in a dynamical spacetime this construction will be due to a certain gauge - your angle will depend on this as well. I wanted to leave out these things for simplicity. Greets –  Robert Filter Jan 18 '11 at 19:34
    
Robert: I'm looking strictly at the internal geodesics (WRT the induced spatial metric) within some timeslice - essentially trapping the problem at a single 'time', for some loose definition of that term. While this doesn't always work, it does generally work for the stationary metrics you're talking about (outside of the wonkier regions, e.g. inside the Schwarzschild horizon!) and lets the problem be phrased as a strictly-spatial problem that IMHO gets more cleanly at the essence of the original question; the (timeslice-internal) geodesics correspond to what we think of as 'lines'. –  Steven Stadnicki Jan 18 '11 at 20:56
    
@Steven: Thank you for your further thoughts. First of all I have to state that the choice of a timeslice would be the gauge I was referring to - hence angles will depend on this choice. Second of all the author is referring to lines of sight - I would interprete these as light geodesics. So, to maybe find a compromise here: One could define such angles more generally but they would depend on the gauge. Is this along your lines of thought? Greets –  Robert Filter Jan 18 '11 at 22:17

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